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(B) The system acts as a physical pendulum oscillating about point $P$. The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I_P}{

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$T = 2\pi \sqrt {\frac{{\sqrt 2 l}}{g}}$

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(B) The system acts as a physical pendulum oscillating about point $P$. The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I_P}{mgd}}$,where $I_P$ is the moment of inertia about the pivot $P$ and $d$ is the distance of the center of mass from $P$.For a circular arc of radius $l$ and angle $2\alpha = 90^\circ$ (i.e.,$\alpha = 45^\circ$),the center of mass is at a distance $d = l \frac{\sin \alpha}{\alpha}$ from the center of the circle. Here $\alpha = \pi/4$ radians,so $d = l \frac{\sin(\pi/4)}{\pi/4} = l \frac{1/\sqrt{2}}{\pi/4} = \frac{2\sqrt{2}l}{\pi}$.The moment of inertia of the arc about its center is $I_{cm} = ml^2$. By the parallel axis theorem,the moment of inertia about the pivot $P$ is $I_P = I_{cm} + md^2 = ml^2 + m(\frac{2\sqrt{2}l}{\pi})^2 = ml^2(1 + \frac{8}{\pi^2})$.However,for a circular arc suspended by strings of length $l$ from $P$,the arc behaves as a rigid body rotating about $P$. The distance of the center of mass from $P$ is $d = l \cos(45^\circ) = l/\sqrt{2}$.The moment of inertia about $P$ is $I_P = ml^2$. Substituting into the formula: $T = 2\pi \sqrt{\frac{ml^2}{mg(l/\sqrt{2})}} = 2\pi \sqrt{\frac{l}{g/\sqrt{2}}} = 2\pi \sqrt{\frac{\sqrt{2}l}{g}}$.

$A$ circular arc of mass $m$ is connected with the help of two massless strings as shown in the figure in a vertical plane. About point $P$,small oscillations are given in the plane of the arc. The time period of the oscillations of $SHM$ will be

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